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Subnet/supernet Question on CCNA Test

Answered Question

Hey,


I had a question on the test yesterday that I am going to try and remember, but may screw it up.  Maybe some others can remember a similar question and get it correct for me.


It have you an address and asked you that you needed to have 80 subnets with 4 usuable host addresses in each subnet with minimal amount of wasted addresses.  Then if gave you some multiple choice options and it looked like, if I can remember the options resembled stuff like this:


192.168.2.4/27 255.255.255.252

etc

etc

etc


Like I said, I am guessing at what the answers are, but I am looking for the theory on how to get this.  Is it just straight up supernetting?


So would it just be like 192.168.2.0/23   or something like that.


Thanks


Gene

Correct Answer by Peter Paluch about 5 years 6 days ago

Hi Gene,


This is straight subnetting. In fact, this subnetting assignment can be be approached from two perspectives.


Assume that the original network is 10.0.64.0/20. Now, if you are asked to create 80 subnets then - because you can't create exactly 80 subnets but only a power of two - what you really want to do is to create 128 subnets. To have 128 subnets, you need to borrow another 7 bits from the current host part which will carry the subnet ID (2^7=128). This will cause the mask to move from /20 to /27, creating the following subnets:


10.0.64.0/27

10.0.64.32/27

10.0.64.64/27

10.0.64.128/27

...

...

10.0.79.128/27

10.0.79.160/27

10.0.79.192/27

10.0.79.224/27


Each of these subnets has a mask of /27 meaning that it holds 32 addresses in total, 30 usable addresses. This approach allowed you to adapt to the required number of subnets but the size of the individual network simply resulted from the changed netmask.


Another approach is to focus not on the number of requested subnets but rather on their size. If a size of 4 usable addresses is required then the real network must have a size of 8 addresses in total (4 usable, 2 for network/bcast, totalling 6 addresses - and again, a size of a network is only a power of two, hence the increase to 8 addresses). A network that requires 8 addresses in total has a distinct netmask of /29. In other words, all subnets would be of the form


10.0.64.0/29

10.0.64.8/29

10.0.64.16/29

...

10.0.79.224/29

10.0.79.232/29

10.0.79.240/29

10.0.79.248/29


Now, because you have selected the mask of /29, you have borrowed additional 9 bits from the former host part of the address (recall that the original netmask was /20). Beause of this, you have created 2^9=512 subnets. So by this approach, you have focused on the requested size of a single network, and hence the number of created subnetwors simply resulted.


In certification exams, the requested size and the number of subnetworks are usually carefully chosen so that using either approach will yield identical results. In this example, requesting 8 subnets for 512 addresses in total would always lead to the selection of the netmask of /23 using either approach (8 subnets requires borrowing three bits, i.e. /20 -> /23; a network of the size of 512 addresses requires 9 bits in the host part, hence the /23 netmask).


I am not sure if this helped so please feel welcome to ask further!


Best regards,

Peter

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Correct Answer
Peter Paluch Sat, 08/11/2012 - 14:24
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  • Cisco Employee,

Hi Gene,


This is straight subnetting. In fact, this subnetting assignment can be be approached from two perspectives.


Assume that the original network is 10.0.64.0/20. Now, if you are asked to create 80 subnets then - because you can't create exactly 80 subnets but only a power of two - what you really want to do is to create 128 subnets. To have 128 subnets, you need to borrow another 7 bits from the current host part which will carry the subnet ID (2^7=128). This will cause the mask to move from /20 to /27, creating the following subnets:


10.0.64.0/27

10.0.64.32/27

10.0.64.64/27

10.0.64.128/27

...

...

10.0.79.128/27

10.0.79.160/27

10.0.79.192/27

10.0.79.224/27


Each of these subnets has a mask of /27 meaning that it holds 32 addresses in total, 30 usable addresses. This approach allowed you to adapt to the required number of subnets but the size of the individual network simply resulted from the changed netmask.


Another approach is to focus not on the number of requested subnets but rather on their size. If a size of 4 usable addresses is required then the real network must have a size of 8 addresses in total (4 usable, 2 for network/bcast, totalling 6 addresses - and again, a size of a network is only a power of two, hence the increase to 8 addresses). A network that requires 8 addresses in total has a distinct netmask of /29. In other words, all subnets would be of the form


10.0.64.0/29

10.0.64.8/29

10.0.64.16/29

...

10.0.79.224/29

10.0.79.232/29

10.0.79.240/29

10.0.79.248/29


Now, because you have selected the mask of /29, you have borrowed additional 9 bits from the former host part of the address (recall that the original netmask was /20). Beause of this, you have created 2^9=512 subnets. So by this approach, you have focused on the requested size of a single network, and hence the number of created subnetwors simply resulted.


In certification exams, the requested size and the number of subnetworks are usually carefully chosen so that using either approach will yield identical results. In this example, requesting 8 subnets for 512 addresses in total would always lead to the selection of the netmask of /23 using either approach (8 subnets requires borrowing three bits, i.e. /20 -> /23; a network of the size of 512 addresses requires 9 bits in the host part, hence the /23 netmask).


I am not sure if this helped so please feel welcome to ask further!


Best regards,

Peter

Excellent.  Thank you, this helped greatly.  I guess you just have to look at it as if you were an ISP. You get this supernet of addresses to dole out.  Then you subnet the address range on the 4th octet to dole those out.


I knew how a supernet worked, I knew how a subnet worked...just never had to put them together.


So, with the /20 then the subnet of that to a /27, would you right that as a 10.0.64.0 255.255.240.224  or you wouldn't even have to express it.  So the answer would be with a x.x.x.x/20  you would need to subnet it with an x.x.x.x/27 to get what you need, so the answer would be the /27. 


Appreciate it. 


Gene

Peter Paluch Mon, 08/13/2012 - 11:45
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  • Cisco Employee,

Hello Gene,


Be careful about using the word "supernetting". I have a feeling you are using it in an incorrect context.


Correctly, the term "supernetting" refers to the process of aggregating a number of networks into a single large network that is larger than the appropriate class. For example, networks 192.168.0.0/24, 192.168.1.0/24, 192.168.2.0/24, and 192.168.3.0/24 can be supernetted into 192.168.0.0/22 - this supernetwork contains four C-class networks. Supernetting is therefore a special case of summarization in which the result has a shorter mask than the appropriate classful mask.


Note that if you aggregate networks, say, 192.168.1.0/26, 192.168.1.64/26, 192.168.1.128/26 and 192.168.1.192/26 into a single network 192.168.1.0/24, you are summarizing these four networks but you are not supernetting here because the result of the summarization does not have its netmask shorter than the appropriate classful mask.


This is arguably a words play but then again, many times, these subtle differences may have quite profound effect.


Best regards,

Peter

Peter,


Thanks, never had it explained that way, which makes sense.  I just figured, given a classful "class B"  and masking on the 3rd octet, like 137.139.0.0 /24,  you would supernet by shortening the mask,  which give you less subnets, but more than 254 nodes per network.  I assumed that was supernetting. With this supernetting or your example above, would that not put ALL the nodes in the same broadcast domain with 192.168.0.0 being the network and 192.168.3.255 being the broadcast


But with the summarization, wouldnt all those addresses (x.x.1.0, x.x.1.64, x.x.1.128, and x.x.1.192) all be a different broadcast domain and this be only a summarization like for a router, or are you really aggreagting them together into one broadcase domain?   


And if that is the case, then it is just by definition only, meaning  creating a reault with a shorter mask than the classful mask (ie 192.168.x.x not /24 anymore but < /24) would be supernetting.   BUT with takein 192.168.x.x/26 addresses and summarizing them with a mask that is =>/24 (its true classful mask) would just be summarization.



Oh, and since you are obviously a genius when it comes to this stuff,  there was another question on the test about which route a packet would take I think using EIGRP.

Packet to 137.139.8.9/30

137.139.8.0/29 = route A

137.139.8.0/28 = route B

137.139.8.0/27 = route C


Or it was something like that.  I figured the route with the most granular mask seeing they all cover the network associated with the 137.139.8.9/30 adddress, OR does this not really matter as you look at Administrative Distance if multiple routing protocols are used, then the route cost and the masks don't matter.  Sorry, off topic, but this question bugged me too.








Thanks for your help


Gene

cadet alain Mon, 08/13/2012 - 12:58
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Hi,

to answer this question about which route will get used the answer is : longest match so answer A in the example you provided.


Regards.


Alain


Don't forget to rate helpful posts.

Peter Paluch Mon, 08/13/2012 - 14:50
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  • Cisco Employee,

Hi Gene,


I just figured, given a classful "class B"  and masking on the 3rd  octet, like 137.139.0.0 /24,  you would supernet by shortening the  mask,  which give you less subnets, but more than 254 nodes per  network.  I assumed that was supernetting.


No, that would not be supernetting. Unless you created a network whose mask would be /15 or less, it is only subnetting into larger networks - but not subnetting.


Think of this: Terms "subnet" and "supernet" are, for simplicity reasons, related to the classful mask. The rule is simple: assume a network A.B.C.D/M. If the M is exactly the classful mask, we say that A.B.C.D is a major network or a classful network. If M is less than the classful mask, we say that A.B.C.D is a supernet. If M is more than the classful mask, we say that A.B.C.D is a subnet. It's as simple as that.


With this supernetting or your example above, would that not put ALL the  nodes in the same broadcast domain with 192.168.0.0 being the network  and 192.168.3.255 being the broadcast


This is a good question. But we have to have a closer look into what the broadcast domain really is. Principially, a broadcast domain is a matter of the Data Link Layer - it is a section of the network that will be flooded with a single broadcast frame once it is sent. Addressing has no influence on the boundaries of a broadcast domain.


So would having the network 192.168.0.0/22 cause all stations in the range 192.168.0.0-192.168.3.255 be put into the same broadcast domain? No. Whether they are or aren't in the same broadcast domain depends exclusively on how they are connected together. If they are placed onto the same Ethernet switched network then yes, they are in the same broadcast domain. If they are split into several Ethernet switched networks interconnected by one or more routers, then no, they will still remain in separate broadcast domains. Pinging the 192.168.3.255 would delivere these so-called directed broadcasts to only a single broadcast domain out of these four - the one that by IP addressing contains the IP address 192.168.3.255.


Regarding that comment about me being a "genius" - no, absolutely not. There are people here on Cisco Support Community who are much more knowledgeable and experienced than I am.


Best regards,

Peter

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